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Qiaochu YuanDec 13 '09 at 16:13

Sorry. In fact, I am using very general framework to revise the constructing injectives in Grothendieck category. I want to test this machinary using this example. However, I suddenly found that I could not prove this isomorphism
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MAJIADec 13 '09 at 16:30

1

Do you mean localized at the point x=0, or localized at the prime ideal (0)?
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Alison MillerDec 13 '09 at 17:07

I am sorry. The multiplicative set should be S=Q[x]-(0)
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MAJIADec 13 '09 at 17:27

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Connected to Scott's answer, you may also like to note that $\mathbb Q[x]/(x) \otimes_{\mathbb Q[x]}\mathbb Q(x) = 0$ since you can always take any basis element $r\otimes f$ and rewrite it as $r\otimes fx = rx\otimes f = 0$. Also, just by the recognition that $\mathbb Q[x]-(0)$ is your multiplicative set, you'll be able to see that the module vanishes; basically by the same argument again: x annihilates all of $\mathbb Q[x]/(x)$, and thus, if $x$ suddenly is invertible, we can take any $r\in\mathbb Q[x]/(x)$ and rewrite its image in the localization to $rx/x = 0/x = 0$.
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Mikael Vejdemo-JohanssonDec 13 '09 at 17:46

2 Answers
2

If you localize Q[x] at zero, you get the field of rational functions Q(x), so your module localizes to a vector space over this field (see Wikipedia). Your module is finitely generated, in fact by a single element, so you get a vector space of dimension zero or one. To decide which is correct, we can use Proposition 2.1 in Eisenbud's Commutative algebra (with a view toward algebraic geometry): If M is finitely generated, then $M[U^{-1}] = 0$ if and only if M is annihilated by an element of U. In this case, x is such an annihilating element, and your localized module is zero. Therefore, your first answer is correct, and your second answer is wrong.